Thread: please help

Two chords KV and QR of circle O are perpendicular at oint P, with PQ = 6 and PR = 8. If the radius of circle O is √65, find KP and PV.

Let $\displaystyle OM\perp RQ$ and $\displaystyle ON\perp KV$.
$\displaystyle RQ=14\Rightarrow MQ=7\Rightarrow ON=MP=MQ-PQ=7-6=1$
In the triangle OMQ: $\displaystyle OM=\sqrt{OQ^2-MQ^2}=\sqrt{65-49}=\sqrt{16}=4\Rightarrow PN=4$
In the triangle ONK: $\displaystyle KN=\sqrt{OK^2-ON^2}=\sqrt{65-1}=8$
Then $\displaystyle KP=KN-PN=8-4=4$ and $\displaystyle PV=KV-KP=16-4=12$

Red dog your a genius any chance you can help with my problem?
http://www.mathhelpforum.com/math-he...arger-one.html

thank you very much red dog!